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a) Ta có:+) \(\frac{12}{16}=\frac{-x}{4}\) <=> 12.4 = 16.(-x)
<=> 48 = -16x
<=> x = 48 : (-16) = -3
+) \(\frac{12}{16}=\frac{21}{y}\) <=> 12y = 21.16
<=> 12y = 336
<=> y = 336 : 12 = 28
+) \(\frac{12}{16}=\frac{z}{-80}\) <=> 12. (-80) = 16z
<=> -960 = 16z
<=> z = -960 : 16 = -60
b) Ta có: \(\frac{x+3}{7+y}=\frac{3}{7}\) <=> (x + 3).7 = 3(7 + y)
<=> 7x + 21 = 21 + 3y
<=> 7x = 3y
<=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x+y}{3+7}=\frac{20}{10}=2\)
=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{7}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.7=14\end{cases}}\)
Vậy ...
Ta có :x/20=13x/260 ; 4/13=80/260 ; 5/13=100/260
vì 4/13<x/20 nên 8/260<13x/260 =>80<13x do đó 6 <x (1)
vì x/20<5/13 nên x/20<100/13=>13x<100 ,do đó x<8 (2)
từ (1) và (2) ta có x =7
vậy x =7
Ta có:\(\frac{4}{13}< \frac{x}{20}< \frac{5}{13}\)
\(\frac{80}{260}< \frac{13x}{260}< \frac{100}{260}\)
\(\Rightarrow80< 13x< 100\)
\(\Rightarrow\)x=7
Vậy x=7
\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)
\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)
\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)
\(=>x=0\)
\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)
\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)
\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)
tu do \(=>x=-7,8;...;0;1;2;3;4\)
a) \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)< \(x\) < \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
Ta có: \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)
=\(\frac{-40}{15}+\frac{21}{15}+\frac{-71}{15}\)
=\(\frac{-90}{15}\)
=\(-6\)
Ta có: \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
=\(\frac{-26}{14}+\frac{19}{14}+\frac{-49}{14}\)
=\(\frac{-56}{14}\)
=\(-4\)
=> \(-6\)< \(x\)<\(-4\)
=> \(x=-5\)
b)\(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)< \(\frac{x}{9}\)<\(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
Ta có: \(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)
=\(\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}+\frac{-11}{31}\right)+\frac{-4}{9}\)
=\(1+\left(-1\right)+\frac{-4}{9}\)
=\(0+\frac{-4}{9}\)
=\(\frac{-4}{9}\)
Ta có: \(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
=\(\frac{-3}{7}+\frac{7}{15}+\frac{-4}{7}+\frac{8}{15}+\frac{2}{3}\)
=\(\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{2}{3}\)
=\(\left(-1\right)+1+\frac{2}{3}\)
=\(0+\frac{2}{3}\)
=\(\frac{2}{3}\)
=> \(\frac{-4}{9}\)< \(\frac{x}{9}\)<\(\frac{2}{3}\)
=
=> \(\frac{-4}{9}\)<\(\frac{x}{9}\)<\(\frac{6}{9}\)
=> \(-4\)< \(x\)<\(6\)
=>\(x\in\left\{-3;-2;-1;0;1;2;3;4;5\right\}\)
Ta có : \(\frac{6}{x}< \frac{x}{3}< \frac{13}{x}\) nên \(\frac{18}{3x}< \frac{x^2}{3x}< \frac{39}{3x}\)
\(\Rightarrow\)18 < x2 < 39. Vậy x2 \(\in\){ 25 ; 36 }. Do x > 0 nên x\(\in\){ 5 ; 6 }
\(\frac{-13}{15}< \frac{y}{5}< \frac{-13}{21}\)
\(\frac{-13}{15}=\frac{-13.7}{15.7}=\frac{-91}{105}\)
\(\frac{y}{5}=\frac{21.y}{5.21}=\frac{21y}{105}\)
\(\frac{-13}{21}=\frac{-13.5}{21.5}=\frac{-65}{105}\)
=>\(\frac{-91}{105}< \frac{21y}{105}< \frac{-65}{105}\)
=> \(-91< 21y< -65\)
=> \(y=-4\)
Vì 21.(-4)=-84 (-91<-84<-65)
=>\(y=-4\)
Chúc bạn học tốt!
\(\frac{148}{37}\)< x < \(\frac{104}{13}\)
4 < x < 8
Mà x thuộc Z
=> x thuộc {5; 6; 7}
\(13\frac{13}{20}< x< 20\frac{31}{20}\Rightarrow\frac{273}{20}< x< \frac{413}{20}\Rightarrow13,65< x< 20,65\)
\(Mà\)\(x\in Z\Rightarrow14\le x\le20\Leftrightarrow x\in\left\{14;15;16;17;18;19;20\right\}\)